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miskamm [114]
3 years ago
14

Why should the lateral sides of the infant's heel be used for capillary puncture?

Biology
1 answer:
bazaltina [42]3 years ago
4 0
Damage to nerve or bone can happen when you puncture infant's extremities to collect capillary blood. The safest region to do this is the lateral side of <span>the infant's heel. With this method, the chance damage to the nerve of bone is much less than the middle of an infant or the finger.</span>
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Answer:

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A signaling pathway that stimulates cellular growth and division is positively regulated by a GTPase encoded by an autosomally l
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Answer:

The correct answer is: a.

Explanation:

  • A Diploid organism possess two copies ( also called alleles) of each  autosomal gene, of which one copy of the gene is obtained from one of its parents and the other copy from its other parent.
  • During the process of replication, one of the alleles of the GTPase encoding gene undergoes a mis-sense mutation.
  • A Mis-sense mutation can be defined as a non-synonymous mutation in which one of the nucleotide in the sequence of the gene gets altered such that it causes a change in the amino acid encoded by the codon (triplet nucleotide message encoding for a single amino acid)  formed by the mutated nucleotide. A single amino acid change in the protein, if occurs in the functional site of the protein, can render a protein non-functional.
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  • A heterozygous cell is one which has one copy of the normal allele and the other copy of the mutated allele of the GTPase encoding gene.
  • As one copy of the normal GTPase allele is present in such an individual, it can produce the normal GTPase protein that can assist in the process of cell growth and division.
  • Hence, the phenotype(expressed character) of  the heterozygous cell will not be affected due to the mutation and it will be able to grow and divide normally.
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3 years ago
In the following diagram, the voltage is 15 volts and the resistance is 3.0 ohms. Use Ohm's law to determine the curre
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Answer:

Explanation:

V = 15 volts

R = 3 ohms

I = ?

Formula

V = I  *  R

15 = I * 3

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5 = AMPS

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3 years ago
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